bulg olym - maths

wiki it man. a lot is given over there

\left(a^{2}-1 \right)x^{2}+1 is a perfect square if and only if x is a term of the sequence defined by x₀=0,x₁=1, x_{k+2}=2ax_{k+1} -x_{k}

now, since m(m+1)(m+2)(m+3)=(m^{2}+3m+1)^{2}, then

n(n+1)^{2}(n+2)^{3}(n+3)^{4}+1=[(n+1)^{2}-1][(n+1)(n+2)(n+3)^{2}]^{2}+1 is a perfect square.

a general result :(\sqrt{a^{2}-1}+1)^{k}=x_{k}\sqrt{a^{2}-1}+y_{k}, then all the solutions of the equation (a^{2}-1)x^{2}+1 are (x_{k},y_{k})

now in the result , with a=n+1 , we get that
n(n+1)(n+2)(n+3)^{2}
is a term of the sequence defined above for k≥0 .

we can also say that remainders of x_k modulo 2n+1 or 2n+3 are 0, 1 or -1 .

hence (n+1)(n+2)(n+3)^{2}\equiv 0,\pm 1mod(2n+1) , then
(2n+2)(2n+4)(2n+6)^{2}\equiv 0,\pm 16mod(2n+1)

now using that 2n+2 , 2n+4 , 2n+6 leave a remainder of 1 , 3 and 5 respectively when divided by 2n+1 , it foloows that 2n+1 divides 75,59 or 91.

using the same argument again , we also get that 2n+3 divided 7, 9 or 25.

the only numbers satisfying both the conditions are n= 1,2,3. which are the required numbers.